OFFSET
1,1
COMMENTS
The coclass cc(M) for one of the fields K with conductor c = a(n) is n-1, and for each field K with conductor c < a(n), the coclass cc(M) is less than n-1. Among the 3-groups M of coclass cc(M)=1, we distinguish the abelian 3-group A=(3,3) by formally putting cc(A)=0, in accordance with the FORMULA. This is a significant difference to quadratic fields, which are firstly uniquely determined by their discriminant, and secondly cannot have an abelian second 3-class group.
LINKS
Siham Aouissi and Daniel C. Mayer, Coclass of the second 3-class group, arXiv:2508.17510 [math.NT], 2025. See p. 11.
Daniel Constantin Mayer, The distribution of second p-class groups on coclass graphs, arXiv:1403.3833 [math.NT], 2014; J. Théor. Nombres Bordeaux 25 (2013), 401-456.
Daniel Constantin Mayer, Magma program "CyclCoClass.m" with endless loop
FORMULA
According to Theorem 3.12 on page 435 of "The distribution of second p-class groups on coclass graphs", the coclass of the non-abelian 3-group M is given by cc(M)+1=log_3(h_3(E_2)), where h_3(E_2) is the second largest 3-class number among the four unramified cyclic cubic extensions E_1,..,E_4 of the cyclic cubic field K, and log_3 denotes the logarithm with respect to the basis 3. An exception is the abelian 3-group A=(3,3) with correct cc(A)=1, where the FORMULA yields cc(A)=0.
EXAMPLE
We have M abelian for c=657=9*73 (two fields in a doublet), cc(M)=1 for c=2439=9*271 (two fields in a doublet), cc(M)=2 for c=7657=13*19*31 (three fields in a quartet), cc(M)=3 for c=41839=7*43*139 (two fields in a quartet), cc(M)=4 for c=231469=7*43*769 (four fields in a quartet). If the conductor c has two prime divisors, then cc(M)=1. For cc(M) > 1, exactly three prime divisors of the conductor c are required.
PROG
(Magma) // See Links section.
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Daniel Constantin Mayer, Jan 15 2025
STATUS
approved